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Chapter 6 test Question 1 Find the reference angle. θ = 240° a. 30° b. 40° c. 50° d. 60° Solution: We know the 360 degrees is the total circle angle. The reference angle is referenced based upon this. So, the angle is -240 you would subtract that from 360. =360-240 = 60 degree Chapter 6 test Question 2 Find the reference angle. θ = 105° a. 105° b. 75° c. 15° d. 195° Solution: First, draw 105 degrees. You should end up in the second quadrant. Now draw a line (terminal line) diagonally from the origin. Your reference angle is the angle that is created between the x-axis and the terminal line. So, to get the reference angle, you must subtract 105 degrees from 180 degrees. = 180 - 105 = 75 Degree Chapter 6 test Question 3 Find the reference angle. Θ=-7π/4 a. π/4 b. 3π/4 c. – π/4 d. -3π/4 Solution: A reference angle is just another way to write the same angle. Because a full circle is 2π, we just add 2π to -7π/4. 8π/4 - 7π/4 = π/4 radian Chapter 6 test Solution: Given tan theta = -sqrt(55)/5 We know from trigonometric formula Sec^2 θ= 1 + tan^2 θ Plug the value of tan theta Sec^2 θ = 1+ (-sqrt(55)/5) = 1+ 55/25 = 80/25 Take Square root both sides. Sec θ = Chapter 6 test Sec θ = 4sqrt(5)/5 or cos θ = 5/ 4sqrt(5) We know, Tan θ = sin θ/cos θ Sin θ = tan θ * cos θ Plug the value of tan theta and cos theta Sin θ = (-sqrt(55)/5) *(5/ 4sqrt(5)) = -sqrt(11)/4 The angle lies in the IV quadrant that is why sin will be negative. So the Sin theta = - (-sqrt(11)/4) = sqrt(11)/4 Option (d) is correct. Chapter 6 test Solution: Given tan theta = -sqrt(114)/19 We know from trigonometric formula Sec^2 θ = 1 + tan^2 θ Plug the value of tan theta Sec^2 θ = 1+ (-sqrt(114)/19)^2 = 1+ 114/361 = 475/361 Take Square root both sides. Sec θ = 5sqrt(19)/19 , the angle lie in the II quadrant that is why sec theta will be negative. Chapter 6 test Sec θ = - 5sqrt(19)/19 Option (a) is correct. c= 173 cm. Solve the triangle for sides a and b, rounding to two decimal places. Question 6 In the triangle below, a. a = 222.61 cm , b = 274.90 cm b. a = 274.90 cm, b = 222.61 cm b. a = 134.45 cm, b = 108.87 cm d. a = 108.87 cm, b = 134.45 cm Solution: Chapter 6 test Apply the formula in above triangle. Sin 39 = adjacent/ hypotenuse Adjacent = sin 39 * hypotenuse Plug the value of hypotenuse (c = 173 cm) and we know the value of sin 39 = 0.6293 Adjacent (a) or (leg 1) = 0.63 * 173 = 108.87 centimeter Now we have two value a = 108.87 cm and c = 173 cm. Apply the Pythagorean Theorem and find the value f b. Adjacent (c) or (leg 2) = Sqrt(c^2 – b^2) = Sqrt (173^2 -108.87^2) = 134.45 cm Chapter 6 test Question 7 Find the measure of angle θ to the nearest tenth of a degree. 10 yd 20 yd a. 63.4° b. 64.3° c. 64.7° d. 65.1° Solution: Apply the formula in above triangle. Tan theta = 20 yd/10 yd = 2 Theta = Arc tan (2) Theta = 63.43 degree Question 8 A person is standing 250 feet from a building. The angle to the top of the building relative to the horizon is 63.1°. How tall is the building? Chapter 6 test a. 491.4 ft b. 492.8 ft c. 494.0 ft d. 494.6 ft Solution: From the question the diagram would be like this. Now, we apply the formula in a triangle ABC. Tan 63.1 = BC/AC Chapter 6 test Plug the value of AC = 250 feet. Height of the building (BC) = 250 feet * Tan 63.1 = 492.8 feet Question 9 y = - 8 sin( x) State the amplitude. a. 8 b. –8 c. 5 d. –5 Solution: Look at the general equation: y = a sin (b (x + c)) + d Where a is the amplitude, b is the frequency, c is the horizontal shift and d is the vertical shift. In a trigonometric function, the period is equal to 2π / b When we compare both the equation we will get the amplitude a Chapter 6 test -8 Question 10: y = 6 sin ( x) State the period. a. π/3 b.2π/3 c. 3π d. 6π Solution: We know that Period = 2π / b ( x) When we compare general equation and y = 6 sin We will get the value of b = 1/3 Then, Period = 2pi/b Chapter 6 test = 2pi/(1/3) = 6pi Solution: In above graph Amplitude = -3 and Period = pi/4 So, pi/4 = 2pi/B, B = 8 W e have the equation in the form y = Asin(Bx) Plug the values. Equation (y) = =-3sin(8x) Option a is correct answer. Chapter 6 test Solution: When we compare above equation to general equation then we will have the amplitude 180 Solution: As we know Period = 2pi/b, When we compare above equation to general equation then we will have the b = pi/9. Period = 2pi/ (pi/9) = 18 Chapter 6 test Solution: Amplitude = 6, period = 2pi/ (pi/20) = 40 Option b have fulfill this criteria Amplitude = 6 and period = 40 Chapter 6 test Solution: Answer of this graph equation y = 5csc (2x) will be option d. Solution: We need to find exact value of Arcsin (- sqrt(2)/2) We know from properties of inverse. Arccos (-x) = pi- Arccos (x) Chapter 6 test We will apply this property here. Arcsin (- sqrt(2)/2) = pi – Arccos(sqrt(2)/2) Now we got the standard value of Arccos(sqrt(2)/2) = pi/4 So, Arcsin (- sqrt(2)/2) = pi – pi/4 = 3pi/4 Solution: We need to find exact value of Arcsin (-1). We know from properties of inverse. Arc sin (-x) = -Arc sin (x) Chapter 6 test We will apply this property here. Arcsin (-1) = - Arcsin (1) Now we got the standard value of Arcsin (1) = pi/2 So - Arcsin (1) = - pi/2 ***End of the solution****